In a class of $9$, there are $4$ students who play soccer. If the teacher chooses $3$ students, what is the probability that all three of them play soccer?
Answer: We can think about this problem as the probability of $3$ events happening. The first event is the teacher choosing one student who plays soccer. The second event is the teacher choosing another student who plays soccer, given that the teacher already chose someone who plays soccer, and so on. The probabilty that the teacher will choose someone who plays soccer is the number of students who play soccer divided by the total number of students: $\dfrac{4} {9}$ Once the teacher's chosen one student, there are only $8$ left. There's also one fewer student who plays soccer, since the teacher isn't going to pick the same student twice. So, the probability that the teacher picks a second student who also plays soccer is $\dfrac{3} {8}$ The probability of the teacher picking two students who play soccer must then be $\dfrac{4} {9} \cdot \dfrac{3} {8}$ We can continue using the same logic for the rest of the students the teacher picks. So, the probability of the teacher picking $3$ students such that all of them play soccer is $\dfrac{4}{9}\cdot\dfrac{3}{8}\cdot\dfrac{2}{7} = \dfrac{1}{21}$